Contents / Calculus / Applications of Integration

Chapter 6

Applications of Integration

Areas, volumes, arc length, surface area, work, hydrostatic force, center of mass, and probability.

Introduction

The definite integral was introduced as area under a curve, but that was only the first thing it measures. The integral is a machine for adding up infinitely many infinitely small contributions, and anything that accumulates — the volume of a vase, the length of a road on a map, the work done by a spring, the push of water on a dam, the balance point of a metal plate, the probability that a phone call arrives in the next two minutes — is a candidate.

Every application in this chapter is built the same way. Cut the quantity into thin slices, approximate the contribution of one slice by something simple (a rectangle, a disk, a cylinder, a constant force), add the approximations to get a Riemann sum, and let the slices shrink. The limit of the Riemann sum is a definite integral by the definition of the integral, and the Fundamental Theorem of Calculus evaluates it. The calculus is identical from problem to problem; what changes is the geometry or physics that tells you what one slice contributes.

Learn the slicing argument once, properly, and you will never need to memorize a formula in this chapter. Every formula below is derived from it, and when a problem does not match any formula on the page — a tank of an unusual shape, a plate of an unusual outline — the derivation is what you fall back on.

6.1Areas between curves

Definition 6.1 (Area between two curves). Let $f$ and $g$ be continuous on $[a,b]$ with $f(x) \ge g(x)$ for all $x$ in $[a,b]$ . The area of the region bounded by $y = f(x)$ above, $y = g(x)$ below, and the vertical lines $x = a$ and $x = b$ is

A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx.

This is a definition rather than a theorem only because "area of a curved region" has no earlier meaning; the integral is what gives it one. The formula is not arbitrary, though. Partition $[a,b]$ into $n$ subintervals of width $\Delta x$ and pick a sample point $x_i^*$ in each. The vertical strip of the region above $[x_{i-1}, x_i]$ is nearly a rectangle of width $\Delta x$ and height $f(x_i^*) - g(x_i^*)$ , so

A \approx \sum_{i=1}^n \bigl[f(x_i^*) - g(x_i^*)\bigr]\,\Delta x,

and the approximation improves as $n$ grows. The right-hand side is a Riemann sum for $f - g$ , so its limit is the integral in the definition. The hypothesis $f \ge g$ is what makes the height $f - g$ non-negative; without it the sum counts some strips with a negative sign and the integral measures a signed area, not an area.

When the curves cross, the integrand $f(x) - g(x)$ changes sign and you must split the interval at the crossing points, taking top minus bottom on each piece separately. The compact way to say this is

A = \int_a^b \bigl|f(x) - g(x)\bigr|\,dx,

but in practice you never integrate an absolute value: you find where $f = g$ , decide which curve is on top on each subinterval, and add the pieces.

Proposition 6.2 (Area with respect to $y$ ). If a region is bounded on the right by $x = h(y)$ , on the left by $x = k(y)$ , and by the horizontal lines $y = c$ and $y = d$ , with $h(y) \ge k(y)$ on $[c,d]$ , then its area is

A = \int_c^d \bigl[h(y) - k(y)\bigr]\,dy.

The proof is the same Riemann-sum argument with horizontal strips of height $\Delta y$ and width $h(y_i^*) - k(y_i^*)$ : right minus left. Integrating in $y$ is the right choice whenever the region's left and right boundaries are each a single curve while its top or bottom boundary changes formula partway across. Choosing the wrong variable does not make the problem impossible, only longer.

Intuition. Think of the region as a stack of vertical matchsticks. Each matchstick starts on the lower curve and ends on the upper one, so its length is top minus bottom, and its sliver of area is that length times its tiny width. Adding the slivers is the integral.

If the matchsticks would have to change which curve they start on halfway across the region, turn them sideways and use horizontal ones instead. That is all "integrating with respect to $y$ " means.

Example 6.3 (Area between a parabola and a line). Find the area of the region enclosed by $y = x^{2}$ and $y = x + 2$ .

Solution.

Find where the curves meet: $x^{2} = x + 2$ gives $x^{2} - x - 2 = (x-2)(x+1) = 0$ , so $x = -1$ and $x = 2$ . These are the limits of integration.
Decide which curve is on top. At $x = 0$ the line gives $2$ and the parabola gives $0$ , so the line is above the parabola on $[-1,2]$ . (Two continuous curves that do not cross inside the interval cannot swap places, so one test point settles it.)
Set up top minus bottom:

A = \int_{-1}^{2}\bigl[(x+2) - x^2\bigr]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.

Evaluate: at $x = 2$ the bracket is $2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}$ , and at $x = -1$ it is $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$ . The difference is $\tfrac{10}{3} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}$ .

Sanity check: the region sits inside the rectangle $[-1,2]\times[0,4]$ of area $12$ and fills something like a third of it. An answer of $4.5$ is plausible.□

Example 6.4 (Curves that cross). Find the area of the region bounded by $y = \sin x$ , $y = \cos x$ , $x = 0$ and $x = \pi/2$ .

Solution.

The curves cross where $\sin x = \cos x$ , that is, $\tan x = 1$ , so at $x = \pi/4$ in this interval.
On $[0, \pi/4]$ the cosine is on top (at $x = 0$ , $\cos 0 = 1 > 0 = \sin 0$ ); on $[\pi/4, \pi/2]$ the sine is on top. Split the integral there:

A = \int_0^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx.

The first integral is $\bigl[\sin x + \cos x\bigr]_0^{\pi/4} = \bigl(\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}\bigr) - (0 + 1) = \sqrt2 - 1$ .
The second is $\bigl[-\cos x - \sin x\bigr]_{\pi/4}^{\pi/2} = (0 - 1) - \bigl(-\tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2}\bigr) = \sqrt2 - 1$ .
Total area: $A = 2\sqrt2 - 2 \approx 0.83$ .

Had we integrated $\cos x - \sin x$ straight across $[0, \pi/2]$ we would have obtained $0$ : the two pieces are congruent and cancel. That is the signed-area trap.□

Example 6.5 (Integrating with respect to $y$ ). Find the area enclosed by the line $y = x - 1$ and the parabola $y^{2} = 2x + 6$ .

Solution.

Solve both for $x$ : the line is $x = y + 1$ and the parabola is $x = \tfrac{1}{2}y^{2} - 3$ .
They meet where $y + 1 = \tfrac12 y^{2} - 3$ , so $y^{2} - 2y - 8 = (y-4)(y+2) = 0$ : at $y = -2$ and $y = 4$ .
For $y$ between $-2$ and $4$ the line lies to the right of the parabola (test $y = 0$ : the line gives $x = 1$ , the parabola $x = -3$ ). Right minus left:

A = \int_{-2}^{4}\left[(y+1) - \left(\tfrac12 y^2 - 3\right)\right]dy = \int_{-2}^{4}\left(-\tfrac12 y^2 + y + 4\right)dy.

An antiderivative is $-\tfrac{y^{3}}{6} + \tfrac{y^{2}}{2} + 4y$ . At $y = 4$ : $-\tfrac{64}{6} + 8 + 16 = \tfrac{40}{3}$ . At $y = -2$ : $\tfrac{8}{6} + 2 - 8 = -\tfrac{14}{3}$ . The area is $\tfrac{40}{3} + \tfrac{14}{3} = 18$ .

Doing this in $x$ would require splitting at $x = -1$ , where the lower boundary switches from the bottom half of the parabola to the line, and integrating square roots. One integral in $y$ replaces two harder integrals in $x$ .□

Example 6.6 (Using symmetry). Find the area enclosed by $y = 2 - x^{2}$ and $y = x^{2}$ .

Solution.

The curves meet where $2 - x^{2} = x^{2}$ , so $x = \pm 1$ , and $2 - x^{2}$ is on top between them.
Both curves are even functions, so the region is symmetric about the $y$ -axis and its area is twice the area of the right half:

A = 2\int_0^1 \bigl[(2 - x^2) - x^2\bigr]\,dx = 2\int_0^1 (2 - 2x^2)\,dx = 2\left[2x - \frac{2x^3}{3}\right]_0^1 = 2\cdot\frac{4}{3} = \frac{8}{3}.

Symmetry halves the arithmetic and, more importantly, halves the chances of a sign slip at the negative endpoint.□

Pitfall. $\int_a^b f(x)\,dx$ is the area between the graph of $f$ and the $x$ -axis only when $f \ge 0$ on $[a,b]$ . The area bounded by $y = x^{3} - x$ and the $x$ -axis on $[-1,1]$ is $\tfrac12$ , but $\int_{-1}^{1}(x^{3} - x)\,dx = 0$ . Whenever a problem says "area", locate the sign changes first.

Exercises for §6.1 →

6.2Volumes by slicing

A solid is harder to measure than a plane region, but the strategy is identical: cut it into thin slabs, approximate each slab by a cylinder in the general sense (a shape with a flat base swept straight upward), and add.

Definition 6.7 (Volume by cross-sections). Let $S$ be a solid lying between the planes $x = a$ and $x = b$ , and let $A(x)$ be the area of the cross-section of $S$ in the plane through $x$ perpendicular to the $x$ -axis. If $A$ is continuous on $[a,b]$ , the volume of $S$ is

V = \int_a^b A(x)\,dx.

The derivation is the area argument raised one dimension. Partition $[a,b]$ and slice $S$ with planes at the partition points into $n$ slabs. The slab between $x_{i-1}$ and $x_i$ is nearly a cylinder with base area $A(x_i^*)$ and height $\Delta x$ , so its volume is nearly $A(x_i^*)\,\Delta x$ , and

V \approx \sum_{i=1}^n A(x_i^*)\,\Delta x \longrightarrow \int_a^b A(x)\,dx.

The formula is exact for the shapes elementary geometry already knows how to measure. A box of base $\ell \times w$ and height $h$ has $A = \ell w$ throughout, so $V = \ell w h$ . A cylinder of radius $r$ has $A = \pi r^{2}$ , so $V = \pi r^{2} h$ . What the integral adds is the ability to handle a cross-section that changes from slice to slice.

Everything hinges on finding $A(x)$ , and $A(x)$ is a plane-geometry problem: what shape is the slice, and what are its dimensions in terms of $x$ ? The three examples below are the three ways the question usually arrives.

Intuition. A loaf of bread is a stack of slices. If every slice has the same shape you can multiply one slice's area by the number of slices; if the slices change shape along the loaf, as in a baguette that tapers, you have to add the slices one at a time, and adding infinitely thin slices is integration.

A solid built on a base region with, say, square cross-sections is a loaf whose slices are squares whose size is dictated by the width of the base at each point.

Example 6.8 (Volume of a sphere). Show that a sphere of radius $r$ has volume $\tfrac43\pi r^{3}$ .

Solution.

Center the sphere at the origin and slice it perpendicular to the $x$ -axis. The slice at position $x$ is a disk whose radius $y$ satisfies $x^{2} + y^{2} = r^{2}$ , so $y = \sqrt{r^{2} - x^{2}}$ .
The cross-sectional area is $A(x) = \pi y^{2} = \pi(r^{2} - x^{2})$ , and the sphere runs from $x = -r$ to $x = r$ .
By symmetry,

V = \int_{-r}^{r}\pi(r^2 - x^2)\,dx = 2\pi\int_0^r (r^2 - x^2)\,dx = 2\pi\left[r^2x - \frac{x^3}{3}\right]_0^r = 2\pi\cdot\frac{2r^3}{3} = \frac{4}{3}\pi r^3.

The familiar formula, which is normally quoted without proof, is a two-line consequence of slicing.□

Example 6.9 (Square cross-sections on a circular base). The base of a solid is the disk $x^{2} + y^{2} \le 1$ . Cross-sections perpendicular to the $x$ -axis are squares with one side lying in the base. Find the volume.

Solution.

At position $x$ , the base extends from $y = -\sqrt{1-x^{2}}$ to $y = \sqrt{1-x^{2}}$ , so the side of the square is $s = 2\sqrt{1-x^{2}}$ .
The cross-sectional area is $A(x) = s^{2} = 4(1-x^{2})$ .
The solid runs from $x = -1$ to $x = 1$ :

V = \int_{-1}^{1} 4(1-x^2)\,dx = 2\int_0^1 4(1 - x^2)\,dx = 8\left[x - \frac{x^3}{3}\right]_0^1 = \frac{16}{3}.

Sanity check: the solid fits inside a cube of side $2$ , of volume $8$ , and its largest square slice has area $4$ ; a volume around $5.3$ is reasonable.□

Example 6.10 (Semicircular cross-sections). The base of a solid is the region bounded by $y = \sqrt{x}$ , the $x$ -axis and $x = 4$ . Cross-sections perpendicular to the $x$ -axis are semicircles whose diameter lies in the base. Find the volume.

Solution.

At position $x$ the base has width $\sqrt{x}$ , which is the diameter of the semicircle, so the radius is $\tfrac12\sqrt{x}$ .
The area of a semicircle of radius $\rho$ is $\tfrac12\pi\rho^{2}$ , so $A(x) = \tfrac12\pi\cdot\tfrac{x}{4} = \tfrac{\pi x}{8}$ .
Integrate from $0$ to $4$ :

V = \int_0^4 \frac{\pi x}{8}\,dx = \frac{\pi}{8}\cdot\frac{16}{2} = \pi.

The only work is step 1: identifying that the width of the base is the diameter, not the radius. Halving it is where the most common error in this kind of problem lives.□

Example 6.11 (Volume of a pyramid). Find the volume of a pyramid whose base is a square of side $b$ and whose height is $h$ .

Solution.

Put the apex at the origin and let the $x$ -axis run down the pyramid's axis of symmetry, so the base sits in the plane $x = h$ .
The cross-section at distance $x$ from the apex is a square, and by similar triangles its side $s$ satisfies $s/b = x/h$ , so $s = bx/h$ and $A(x) = b^{2}x^{2}/h^{2}$ .
Then

V = \int_0^h \frac{b^2}{h^2}x^2\,dx = \frac{b^2}{h^2}\cdot\frac{h^3}{3} = \frac{1}{3}b^2 h.

This is the classical "one third of base times height", and the same calculation works for a cone or for any pyramid: the cross-sectional area scales as $x^{2}/h^{2}$ , and $\int_0^h x^{2}\,dx = h^{3}/3$ supplies the $\tfrac13$ .□

Exercises for §6.2 →

6.3Solids of revolution: disks and washers

A solid of revolution is what you get by rotating a plane region about a line in its plane. Because the region is being spun, every cross-section perpendicular to the axis is a circle or a ring, and slicing the solid perpendicular to the axis turns the general formula $V = \int A\,dx$ into two specific ones.

Theorem 6.12 (Disk method). If the region under $y = f(x)$ , above the $x$ -axis, from $x = a$ to $x = b$ is rotated about the $x$ -axis, the resulting solid has volume

V = \int_a^b \pi\bigl[f(x)\bigr]^2\,dx.

Proof. Slice the solid with the plane through $x$ perpendicular to the $x$ -axis. The region's vertical segment at $x$ runs from the axis to height $f(x)$ ; when rotated it sweeps out a disk of radius $f(x)$ , so $A(x) = \pi[f(x)]^2$ . The volume-by-slicing formula finishes the proof. Written out, the Riemann sum being taken is $\sum \pi[f(x_i^*)]^2\,\Delta x$ : the sum of the volumes of thin coins of radius $f(x_i^*)$ and thickness $\Delta x$ .∎

Theorem 6.13 (Washer method). If the region between $y = f(x)$ (outer) and $y = g(x)$ (inner), with $f(x) \ge g(x) \ge 0$ on $[a,b]$ , is rotated about the $x$ -axis, the resulting solid has volume

V = \int_a^b \pi\left(\bigl[f(x)\bigr]^2 - \bigl[g(x)\bigr]^2\right)dx.

Proof. Now the vertical segment at $x$ runs from height $g(x)$ to height $f(x)$ , so rotating it produces a ring (a washer) with outer radius $R = f(x)$ and inner radius $r = g(x)$ . Its area is $\pi R^{2} - \pi r^{2}$ , and slicing does the rest.∎

Three variations cover every disk-and-washer problem you will meet, and none of them needs a new formula, only the same reasoning applied to a different picture.

Rotation about the $y$ -axis: slice perpendicular to the $y$ -axis, so the slices are indexed by $y$ . Describe the region's boundary curves as $x = h(y)$ , and the radii are horizontal distances:

V = \int_c^d \pi\bigl[h(y)\bigr]^2\,dy,

or the washer version with two curves.

Rotation about a horizontal line $y = c$ : slice perpendicular to that line, again with vertical slices indexed by $x$ , but measure radii from the line, not from the $x$ -axis. A curve $y = f(x)$ sits at distance $|f(x) - c|$ from the line. Be careful about which curve is farther from the axis: rotating about a line above the region makes the lower curve the outer radius.

Rotation about a vertical line $x = c$ : slice perpendicular to it, indexed by $y$ , with radii $|h(y) - c|$ .

In every case the rule is the same one sentence: integrate along the axis of rotation, with each radius the distance from the axis to a boundary curve.

Intuition. A potter's wheel spins a lump of clay, and the profile the potter's hand traces becomes the outline of a vase. To measure the vase's volume, slice it horizontally into thin coins; each coin's volume is $\pi(\text{radius})^2$ times its thickness, and the radius at each height is read straight off the profile. If the vase is hollow, each coin is a ring and you subtract the hole.

Example 6.14 (Disk method about the $x$ -axis). Find the volume of the solid obtained by rotating the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$ -axis.

Solution.

The slice at $x$ is a disk of radius $\sqrt{x}$ , so $A(x) = \pi(\sqrt{x})^2 = \pi x$ .
Integrate:

V = \int_0^4 \pi x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi.

Sanity check: the solid fits inside a cylinder of radius $2$ and length $4$ , of volume $16\pi$ , and the paraboloid fills exactly half of its circumscribed cylinder. Half of $16\pi$ is $8\pi$ .□

Example 6.15 (Washer method about the $x$ -axis). The region enclosed by $y = x$ and $y = x^{2}$ is rotated about the $x$ -axis. Find the volume.

Solution.

The curves meet at $x = 0$ and $x = 1$ , and on $[0,1]$ the line $y = x$ is above the parabola.
The outer radius is $R = x$ and the inner radius is $r = x^{2}$ , so the washer at $x$ has area $\pi(x^{2} - x^{4})$ .
Then

V = \pi\int_0^1 (x^2 - x^4)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac13 - \frac15\right) = \frac{2\pi}{15}.

Note that $R^{2} - r^{2}$ is not $(R - r)^2$ : the washer's area is the difference of two disk areas, and $\pi(x - x^{2})^2$ would be wrong.□

Example 6.16 (Rotation about the $y$ -axis). The region bounded by $y = x^{3}$ , $y = 8$ and $x = 0$ is rotated about the $y$ -axis. Find the volume.

Solution.

Because the axis is vertical, slice perpendicular to it: horizontal slices, indexed by $y$ from $0$ to $8$ .
The slice at height $y$ extends from $x = 0$ to the curve, so its radius is the $x$ -value on the curve: $x = y^{1/3}$ .
Each slice is a disk of area $\pi y^{2/3}$ , so

V = \int_0^8 \pi y^{2/3}\,dy = \pi\left[\frac{3}{5}y^{5/3}\right]_0^8 = \frac{3\pi}{5}\cdot 32 = \frac{96\pi}{5}.

Sanity check: the solid is inside a cylinder of radius $2$ and height $8$ , of volume $32\pi$ , and $\tfrac{96}{5}\pi \approx 0.6 \times 32\pi$ is a plausible fraction for a solid that hugs the cylinder near the top.□

Example 6.17 (Rotation about the line $y = 2$ ). The region enclosed by $y = x$ and $y = x^{2}$ is rotated about the line $y = 2$ . Find the volume.

Solution.

The axis is horizontal, so slice vertically and index by $x$ from $0$ to $1$ .
The axis lies above the region. The parabola $y = x^{2}$ is the curve farther from $y = 2$ , so it gives the outer radius $R = 2 - x^{2}$ ; the line gives the inner radius $r = 2 - x$ .
Set up the washers:

V = \pi\int_0^1\left[(2 - x^2)^2 - (2 - x)^2\right]dx = \pi\int_0^1\left(x^4 - 5x^2 + 4x\right)dx.

Evaluate: $\pi\left[\tfrac{x^{5}}{5} - \tfrac{5x^{3}}{3} + 2x^{2}\right]_0^1 = \pi\left(\tfrac15 - \tfrac53 + 2\right) = \pi\cdot\tfrac{3 - 25 + 30}{15} = \tfrac{8\pi}{15}$ .

Compare with the rotation of the same region about the $x$ -axis, which gave $\tfrac{2\pi}{15}$ . The region is the same size, but rotating it about a more distant axis sweeps out a bigger solid.□

Example 6.18 (Rotation about the line $x = -1$ ). The region enclosed by $y = x$ and $y = x^{2}$ is rotated about the line $x = -1$ . Find the volume with washers.

Solution.

The axis is vertical, so slice horizontally and index by $y$ from $0$ to $1$ .
At height $y$ the region runs from the line ( $x = y$ ) on the left to the parabola ( $x = \sqrt{y}$ ) on the right. The axis $x = -1$ is to the left of both, so the parabola is the outer curve: $R = \sqrt{y} - (-1) = \sqrt{y} + 1$ , and $r = y + 1$ .
Washers:

V = \pi\int_0^1\left[(\sqrt{y} + 1)^2 - (y + 1)^2\right]dy = \pi\int_0^1\left(y + 2\sqrt{y} + 1 - y^2 - 2y - 1\right)dy = \pi\int_0^1\left(2\sqrt{y} - y - y^2\right)dy.

Evaluate: $\pi\left[\tfrac{4}{3}y^{3/2} - \tfrac{y^{2}}{2} - \tfrac{y^{3}}{3}\right]_0^1 = \pi\left(\tfrac43 - \tfrac12 - \tfrac13\right) = \tfrac{\pi}{2}$ .

We will redo this with cylindrical shells in the next section and get the same $\tfrac{\pi}{2}$ with less algebra.□

Pitfall. Radii are distances to the axis of rotation, not $y$ -values. For an axis $y = c$ the radius of the curve $y = f(x)$ is $|f(x) - c|$ , and which of two curves is "outer" depends on where the axis is, not on which curve is higher. Draw the axis on your sketch before writing a single radius.

Exercises for §6.3 →

6.4Volumes by cylindrical shells

Try to rotate the region under $y = 2x^{2} - x^{3}$ , $0 \le x \le 2$ , about the $y$ -axis with washers. You need the boundary as $x$ in terms of $y$ , which means solving a cubic, and the inner and outer radii come from different branches of the solution. The shell method sidesteps this by slicing parallel to the axis instead of perpendicular to it.

Theorem 6.19 (Shell method). If the region under $y = f(x)$ , above the $x$ -axis, from $x = a$ to $x = b$ (with $0 \le a < b$ ) is rotated about the $y$ -axis, the resulting solid has volume

V = \int_a^b 2\pi x\, f(x)\,dx.

Proof. Partition $[a,b]$ and let $\bar x_i$ be the midpoint of $[x_{i-1}, x_i]$ . The rectangle of width $\Delta x$ and height $f(\bar x_i)$ standing on that subinterval, when rotated about the $y$ -axis, sweeps out a cylindrical shell: a solid cylinder of radius $x_i$ and height $f(\bar x_i)$ with a cylinder of radius $x_{i-1}$ removed. Its volume is

\pi x_i^2 f(\bar x_i) - \pi x_{i-1}^2 f(\bar x_i) = \pi\bigl(x_i + x_{i-1}\bigr)\bigl(x_i - x_{i-1}\bigr) f(\bar x_i) = 2\pi\,\bar x_i\, f(\bar x_i)\,\Delta x,

using $x_i + x_{i-1} = 2\bar x_i$ . Adding the shells,

V \approx \sum_{i=1}^n 2\pi\,\bar x_i\, f(\bar x_i)\,\Delta x,

which is a Riemann sum for $2\pi x f(x)$ , and its limit as $n \to \infty$ is the integral.∎

The formula is easier to remember from the shape of one shell than from the symbols. Cut a thin shell along a vertical line and unroll it: you get a rectangular slab of length equal to the circumference $2\pi x$ , height $f(x)$ , and thickness $dx$ . Its volume is (circumference)(height)(thickness), and that is the integrand. In words,

V = \int_a^b 2\pi\,(\text{radius})\,(\text{height})\,(\text{thickness}),

and every shell problem consists of identifying those three quantities. For a region between two curves $f \ge g$ , the height is $f(x) - g(x)$ . For rotation about a vertical line $x = c$ with the region to its right, the radius is $x - c$ . For rotation about the $x$ -axis, slice horizontally: the radius is $y$ , the height is the horizontal extent of the region at height $y$ , and you integrate in $y$ .

Shells and washers always agree, so the choice is purely one of convenience. Use shells when the region is naturally described by vertical strips but the axis is vertical, or by horizontal strips but the axis is horizontal, or whenever the washer method would force you to invert a function or split the integral. Use washers when the strips are already perpendicular to the axis.

Intuition. An onion and a loaf of bread are two ways to cut the same solid. Washers slice it like bread, perpendicular to the axis, and each slice is a disk or ring. Shells peel it like an onion, in layers wrapped around the axis, and each layer is a thin hollow cylinder. When a region's description makes one cut easy and the other awkward, take the easy cut; the volume does not care which one you use.

Example 6.20 (Shells about the $y$ -axis). Find the volume of the solid obtained by rotating the region under $y = x^{2}$ from $x = 0$ to $x = 2$ about the $y$ -axis.

Solution.

The shell at $x$ has radius $x$ , height $x^{2}$ and thickness $dx$ .
Set up and evaluate:

V = \int_0^2 2\pi x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 8\pi.

As a check, do it with washers: slice horizontally, $y$ from $0$ to $4$ , outer radius $2$ and inner radius $\sqrt{y}$ , giving $\pi\int_0^4(4 - y)\,dy = \pi(16 - 8) = 8\pi$ . Same answer, more setup.

□

Example 6.21 (When washers fail and shells succeed). The region under $y = 2x^{2} - x^{3}$ from $x = 0$ to $x = 2$ is rotated about the $y$ -axis. Find the volume.

Solution.

Washers would require expressing the curve as $x$ in terms of $y$ , which means solving a cubic; the horizontal slice at height $y$ also hits the curve twice, at two different roots. Shells avoid all of it.
The shell at $x$ has radius $x$ and height $2x^{2} - x^{3}$ :

V = \int_0^2 2\pi x\left(2x^2 - x^3\right)dx = 2\pi\int_0^2\left(2x^3 - x^4\right)dx = 2\pi\left[\frac{x^4}{2} - \frac{x^5}{5}\right]_0^2 = 2\pi\left(8 - \frac{32}{5}\right) = \frac{16\pi}{5}.

Sanity check: the region's maximum height is at $x = 4/3$ where $y = 32/27 \approx 1.19$ , so the solid fits in a cylinder of radius $2$ and height $1.19$ , of volume roughly $15$ . Our $16\pi/5 \approx 10$ is comfortably inside.□

Example 6.22 (Shells about the line $x = -1$ ). Redo the rotation of the region between $y = x$ and $y = x^{2}$ about the line $x = -1$ using shells.

Solution.

Use vertical strips, $x$ from $0$ to $1$ . The strip at $x$ is at distance $x - (-1) = x + 1$ from the axis, and its height is $x - x^{2}$ .
Shells:

V = \int_0^1 2\pi(x+1)\left(x - x^2\right)dx = 2\pi\int_0^1\left(x - x^3\right)dx = 2\pi\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = 2\pi\cdot\frac14 = \frac{\pi}{2},

where the middle step used $(x+1)(x - x^{2}) = x^{2} - x^{3} + x - x^{2} = x - x^{3}$ .

This agrees with the washer computation from the previous section, and the integrand here is a polynomial with no square roots.

□

Example 6.23 (Shells about the $x$ -axis). Use shells to find the volume when the region bounded by $y = \sqrt{x}$ , $y = 0$ and $x = 4$ is rotated about the $x$ -axis.

Solution.

For a horizontal axis, shells are horizontal strips: index by $y$ from $0$ to $2$ (the curve reaches $y = 2$ at $x = 4$ ).
The strip at height $y$ runs horizontally from the curve, where $x = y^{2}$ , to $x = 4$ , so its length is $4 - y^{2}$ . Its distance from the $x$ -axis is $y$ .
Shells:

V = \int_0^2 2\pi y\left(4 - y^2\right)dy = 2\pi\left[2y^2 - \frac{y^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi.

This matches the disk computation earlier in the chapter, as it must. Here the disk method was easier; the example shows that shells work in either orientation.□

Pitfall. The radius of a shell is the distance from the strip to the axis of rotation, which is $x$ only when the axis is the $y$ -axis. For the axis $x = 3$ and a region to its left, the radius is $3 - x$ , and reversing the sign gives a negative "volume" that should stop you immediately. The height is the length of the strip, never the value $f(x)$ alone when the region has a lower boundary above the axis.

Exercises for §6.4 →

6.5Work

In physics, the work done by a constant force $F$ moving an object a distance $d$ in the direction of the force is $W = Fd$ . In SI units force is in newtons and distance in meters, and work is in joules ( $1\ \text{J} = 1\ \text{N}\cdot\text{m}$ ); in the US customary system force is in pounds, distance in feet, and work is in foot-pounds. The definition breaks the moment the force varies along the way — stretching a spring, lifting a rope that gets lighter as it comes up, or pumping water that has to rise farther the deeper it starts.

Definition 6.24 (Work done by a variable force). If an object moves along the $x$ -axis from $x = a$ to $x = b$ under a force $F(x)$ directed along the axis, the work done is

W = \int_a^b F(x)\,dx.

The formula comes from the same slicing. Partition $[a,b]$ ; on a short subinterval $[x_{i-1}, x_i]$ the force is nearly constant at $F(x_i^*)$ , so the work over that step is nearly $F(x_i^*)\,\Delta x$ , and the total is nearly $\sum F(x_i^*)\,\Delta x$ . Letting the steps shrink gives the integral. Notice that $\int_a^b F\,dx$ is the area under the force-displacement graph, which is why a work problem is often solved fastest by sketching that graph.

Theorem 6.25 (Hooke's Law). The force required to hold a spring stretched (or compressed) $x$ units beyond its natural length is $F(x) = kx$ , where $k > 0$ is the spring constant. Consequently the work needed to stretch the spring from $x = a$ to $x = b$ beyond its natural length is

W = \int_a^b kx\,dx = \frac{k}{2}\left(b^2 - a^2\right).

The law itself is an empirical fact valid for moderate extensions; the work formula is the definition applied to it. The variable $x$ is displacement from the natural length, not the total length of the spring, and $k$ is usually not given directly but must be deduced from one measurement of force at one extension.

Method 6.26 (Cable and rope problems). When a hanging cable of weight density $w$ (weight per unit length) is hauled up over a height $h$ , slice the cable rather than the motion. The piece of cable at initial depth $x$ below the top has weight $w\,dx$ and must be raised a distance $x$ , so

W = \int_0^h w\,x\,dx = \frac{w h^2}{2},

plus, if a load hangs from the end, the load's weight times $h$ .

Method 6.27 (Pumping problems). To find the work needed to pump a liquid out of a tank, slice the liquid into thin horizontal layers, because every drop in one layer travels the same distance. Choose a coordinate $y$ along the vertical, and for the layer at position $y$ determine three things: its cross-sectional area $A(y)$ from the tank's shape, so that its volume is $A(y)\,dy$ ; its weight $\rho g\,A(y)\,dy$ , where $\rho$ is the density of the liquid; and the distance $d(y)$ it must be lifted to the outlet. Then

W = \int_{y_1}^{y_2} \rho g\,A(y)\,d(y)\,dy,

with $y_1$ and $y_2$ the bottom and top of the liquid. For water, $\rho g = 9800\ \text{N/m}^3$ in SI units, and the weight density is $62.5\ \text{lb/ft}^3$ in US units (in which case omit the $g$ ).

The two hard parts of any pumping problem are geometry and bookkeeping. The geometry is $A(y)$ : for a cylinder it is constant, for a cone it comes from similar triangles, for a sphere from Pythagoras. The bookkeeping is $d(y)$ : it depends on where you put the origin and on whether the outlet is at the top of the tank or above it, and a sign error here is the usual failure mode. Write the two functions down separately and check each against a specific layer (the top one and the bottom one) before integrating.

Intuition. Pushing a box across a floor at constant force is just force times distance. Pulling a bungee cord is different: the first few centimeters are easy and the last few are hard, so no single force describes the job. Break the pull into tiny steps, treat the force as constant during each step, and add up force times step length. That sum is the integral, and the same idea handles a rope that gets lighter as it comes up and water that has farther to travel the deeper it sits.

Example 6.28 (Work to stretch a spring). A spring has natural length $10$ cm and spring constant $k = 40$ N/m. Find the work needed to stretch it from a length of $12$ cm to a length of $15$ cm.

Solution.

Measure $x$ in meters from the natural length: a length of $12$ cm is $x = 0.02$ and a length of $15$ cm is $x = 0.05$ .
Hooke's Law gives $F(x) = 40x$ , so

W = \int_{0.02}^{0.05} 40x\,dx = 20\left[x^2\right]_{0.02}^{0.05} = 20(0.0025 - 0.0004) = 0.042\ \text{J}.

Sanity check: the force ranges from $0.8$ N to $2$ N over a displacement of $0.03$ m, so the work should be between $0.024$ J and $0.06$ J. It is.

□

Example 6.29 (Hauling up a cable). A $200$ -lb cable is $100$ ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building?

Solution.

The cable weighs $2$ lb/ft. Slice it: the piece at depth $x$ ft below the top, of length $dx$ , weighs $2\,dx$ lb and must rise $x$ ft.
The work on that piece is $2x\,dx$ , and adding the pieces,

W = \int_0^{100} 2x\,dx = \left[x^2\right]_0^{100} = 10{,}000\ \text{ft-lb}.

Sanity check: if the whole $200$ -lb cable were lifted through its center of mass at $50$ ft, the work would be $200 \times 50 = 10{,}000$ ft-lb. The center of mass shortcut agrees, and it always does when the density is uniform, as we will see in the section on centers of mass.

□

Example 6.30 (Pumping water from a conical tank). A tank has the shape of an inverted circular cone with height $10$ m and top radius $4$ m. It is filled with water to a depth of $8$ m. Find the work required to empty the tank by pumping all the water to the top of the tank. (Take $\rho g = 9800\ \text{N/m}^3$ .)

Solution.

Put the origin at the apex with $y$ measured upward, so the water occupies $0 \le y \le 8$ and the outlet is at $y = 10$ .
Cross-sectional area: by similar triangles the radius at height $y$ is $r = \tfrac{4}{10}y = \tfrac{2}{5}y$ , so $A(y) = \pi r^{2} = \tfrac{4\pi}{25}y^{2}$ .
Lift distance: the layer at height $y$ must rise to $y = 10$ , so $d(y) = 10 - y$ . Check the extremes: the top layer ( $y = 8$ ) rises $2$ m and the bottom layer rises $10$ m. Correct.
Assemble:

W = \int_0^8 9800\cdot\frac{4\pi}{25}y^2\,(10 - y)\,dy = 1568\pi\int_0^8\left(10y^2 - y^3\right)dy = 1568\pi\left[\frac{10y^3}{3} - \frac{y^4}{4}\right]_0^8.

The bracket is $\tfrac{5120}{3} - 1024 = \tfrac{2048}{3}$ , so $W = 1568\pi\cdot\tfrac{2048}{3} \approx 3.36\times10^6\ \text{J}$ .

The water near the apex is a small volume lifted a long way and the water near the surface is a large volume lifted a short way; the integrand $y^{2}(10 - y)$ records exactly that trade-off.□

Example 6.31 (Pumping water from a hemispherical bowl). A hemispherical bowl of radius $3$ m is full of water, with its flat rim at ground level. Find the work required to pump all the water up to the rim.

Solution.

Put the origin at the center of the rim and measure $y$ downward, so the water occupies $0 \le y \le 3$ and a layer at depth $y$ must rise exactly $y$ : $d(y) = y$ .
The horizontal cross-section at depth $y$ is a circle whose radius $r$ satisfies $r^{2} + y^{2} = 9$ , so $A(y) = \pi(9 - y^{2})$ .
Assemble:

W = \int_0^3 9800\,\pi(9 - y^2)\,y\,dy = 9800\pi\left[\frac{9y^2}{2} - \frac{y^4}{4}\right]_0^3 = 9800\pi\left(\frac{81}{2} - \frac{81}{4}\right) = 9800\pi\cdot\frac{81}{4} = 198{,}450\pi \approx 6.23\times10^5\ \text{J}.

Sanity check: the water's weight is $9800\cdot\tfrac{2}{3}\pi\cdot 27 \approx 5.5\times10^5$ N, and the whole mass rises on average a bit over $1$ m (its center of mass is $\tfrac{3}{8}\cdot 3 = 1.125$ m below the rim). Weight times $1.125$ m is $6.2\times10^5$ J, in agreement.□

Example 6.32 (Work against gravity). Newton's law of gravitation says the Earth attracts a body of mass $m$ at distance $r$ from the Earth's center with force $F = GMm/r^{2}$ , which at the surface $r = R$ equals $mg$ . How much work is required to lift a $1000$ kg satellite from the surface to a height $R$ above it, one Earth radius up? Use $R = 6.37\times10^6$ m and $g = 9.8\ \text{m/s}^2$ .

Solution.

Since $GMm/R^{2} = mg$ , write the force as $F(r) = mgR^{2}/r^{2}$ ; the force is not constant, so a naive $mgh$ is wrong.
The work is

W = \int_R^{2R}\frac{mgR^2}{r^2}\,dr = mgR^2\left[-\frac{1}{r}\right]_R^{2R} = mgR^2\left(\frac{1}{R} - \frac{1}{2R}\right) = \frac{mgR}{2}.

Numerically, $W = \tfrac12\cdot 1000\cdot 9.8\cdot 6.37\times10^6 \approx 3.12\times10^{10}$ J.

Using $mgh$ with $h = R$ would give twice this, because it assumes the surface value of gravity all the way up, while in fact the pull has fallen to a quarter of its surface strength by the time the satellite arrives.□

Pitfall. In pumping problems the distance a layer travels is measured to the outlet, not to the water's surface, and the limits of integration cover the water, not the tank. A tank filled to $8$ m out of $10$ m has layers from $y = 0$ to $y = 8$ that each rise $10 - y$ . Mixing those two numbers up is the most common error in this section.

Exercises for §6.5 →

6.6Average value of a function

The average of finitely many numbers is their sum divided by how many there are. A function on an interval takes infinitely many values, and the integral replaces the sum.

Definition 6.33 (Average value of a function). The average value of a continuous function $f$ on $[a,b]$ is

f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx.

To see that this is the right definition, sample $f$ at $n$ equally spaced points $x_1, \dots, x_n$ and average the samples. With $\Delta x = (b-a)/n$ , so that $n = (b-a)/\Delta x$ ,

\frac{f(x_1) + \cdots + f(x_n)}{n} = \frac{1}{b-a}\sum_{i=1}^n f(x_i)\,\Delta x,

and as $n \to \infty$ the Riemann sum on the right tends to the integral. Geometrically, $f_{\text{avg}}$ is the height of the rectangle on $[a,b]$ whose area equals the area under the curve: multiply the definition through by $b - a$ to see that $f_{\text{avg}}(b-a) = \int_a^b f\,dx$ .

Theorem 6.34 (Mean Value Theorem for Integrals). If $f$ is continuous on $[a,b]$ , then there is a number $c$ in $[a,b]$ such that

f(c) = f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx, \qquad\text{that is,}\qquad \int_a^b f(x)\,dx = f(c)\,(b - a).

Proof. By the Extreme Value Theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$ . Since $m \le f(x) \le M$ for all $x$ in the interval, integrating gives $m(b-a) \le \int_a^b f\,dx \le M(b-a)$ , and dividing by $b - a$ ,

m \le f_{\text{avg}} \le M.

So $f_{\text{avg}}$ lies between two values that $f$ takes. By the Intermediate Value Theorem, $f$ takes every value between $m$ and $M$ , so $f(c) = f_{\text{avg}}$ for some $c$ in $[a,b]$ .∎

Continuity is doing all the work: it supplies the extreme values and the intermediate values. For a step function that equals $0$ on the left half of the interval and $1$ on the right half, the average value is $\tfrac12$ , and no point achieves it.

An alternative proof applies the ordinary Mean Value Theorem to $F(x) = \int_a^x f(t)\,dt$ : by the Fundamental Theorem of Calculus $F' = f$ , and the MVT gives a $c$ with $F(b) - F(a) = F'(c)(b - a)$ , which is the same statement. The two proofs are two views of the same fact.

Intuition. A day's temperature rises and falls. The average temperature is not the midpoint of the high and the low; a day that is cold for twenty hours and hot for four has a low average even if the extremes are symmetric. The integral weights every minute equally, so the average value is the constant temperature that would deliver the same total heating over the day. The Mean Value Theorem for Integrals says that at some instant the thermometer actually read that average — it cannot jump over it.

Example 6.35 (Average temperature). The temperature in a city is modeled by $T(t) = 20 + 10\sin\!\left(\tfrac{\pi}{12}t\right)$ degrees Celsius, where $t$ is hours after midnight. Find the average temperature between $6$ am and $6$ pm.

Solution.

The interval is $[6,18]$ , of length $12$ :

T_{\text{avg}} = \frac{1}{12}\int_6^{18}\left(20 + 10\sin\frac{\pi t}{12}\right)dt = \frac{1}{12}\left[20t - \frac{120}{\pi}\cos\frac{\pi t}{12}\right]_6^{18}.

At $t = 18$ : $360 - \tfrac{120}{\pi}\cos\tfrac{3\pi}{2} = 360$ . At $t = 6$ : $120 - \tfrac{120}{\pi}\cos\tfrac{\pi}{2} = 120$ .
$T_{\text{avg}} = \tfrac{240}{12} = 20^\circ\text{C}$ .

The sine term contributes nothing because over $[6,18]$ it spends as much time above zero as below, in mirror image. The temperature peaks at $30^\circ$ at noon, but the average over the daytime hours is the baseline.□

Example 6.36 (Finding the point $c$ ). Find the average value of $f(x) = 1 + x^{2}$ on $[-1,2]$ , and find all $c$ in the interval at which $f(c)$ equals it.

Solution.

Average value:

f_{\text{avg}} = \frac{1}{2-(-1)}\int_{-1}^{2}(1 + x^2)\,dx = \frac13\left[x + \frac{x^3}{3}\right]_{-1}^{2} = \frac13\left[\left(2 + \frac83\right) - \left(-1 - \frac13\right)\right] = \frac13\cdot 6 = 2.

Solve $f(c) = 2$ : $1 + c^{2} = 2$ gives $c = \pm 1$ , and both lie in $[-1,2]$ .

The Mean Value Theorem for Integrals promised at least one such $c$ ; here there are two. Geometrically, the rectangle of height $2$ over $[-1,2]$ has area $6$ , the same as the area under the parabola.□

Example 6.37 (Average of a reciprocal). Find the average value of $f(x) = 1/x$ on $[1, e]$ .

Solution.

Apply the definition:

f_{\text{avg}} = \frac{1}{e - 1}\int_1^e\frac{dx}{x} = \frac{1}{e-1}\Bigl[\ln x\Bigr]_1^e = \frac{1}{e - 1}.

Sanity check: $f$ decreases from $1$ to $1/e \approx 0.37$ , and $1/(e-1) \approx 0.58$ lies between them, as the proof of the Mean Value Theorem for Integrals says it must.

□

Exercises for §6.6 →

6.7Arc length

A curve has a length, but a curve is not a line segment and no ruler measures it directly. The definition is the natural one: approximate the curve by a polygon, measure the polygon, and refine.

Let $f$ have a continuous derivative on $[a,b]$ (such an $f$ is called smooth, and the hypothesis is essential below). Partition $[a,b]$ and let $P_i = (x_i, f(x_i))$ . The polygon with vertices $P_0, P_1, \dots, P_n$ approximates the curve, and its length is $\sum |P_{i-1}P_i|$ . The length $L$ of the curve is defined to be the limit of these polygonal lengths as the mesh of the partition goes to zero, provided the limit exists.

Theorem 6.38 (Arc length formula). If $f'$ is continuous on $[a,b]$ , the length of the curve $y = f(x)$ , $a \le x \le b$ , is

L = \int_a^b\sqrt{1 + \bigl[f'(x)\bigr]^2}\,dx.

Proof. Let $\Delta y_i = f(x_i) - f(x_{i-1})$ . The $i$ -th segment of the polygon has length

|P_{i-1}P_i| = \sqrt{(\Delta x)^2 + (\Delta y_i)^2}.

By the Mean Value Theorem applied to $f$ on $[x_{i-1}, x_i]$ , there is an $x_i^*$ in that subinterval with $\Delta y_i = f'(x_i^*)\,\Delta x$ . Substituting,

|P_{i-1}P_i| = \sqrt{(\Delta x)^2 + [f'(x_i^*)]^2(\Delta x)^2} = \sqrt{1 + [f'(x_i^*)]^2}\;\Delta x,

so the polygon's length is $\sum_{i=1}^n\sqrt{1 + [f'(x_i^*)]^2}\,\Delta x$ , a Riemann sum for the continuous function $\sqrt{1 + (f')^2}$ . As the mesh goes to zero the sum converges to the integral.∎

The proof shows where the hypothesis enters: without a continuous $f'$ the Mean Value Theorem step fails and the Riemann sums need not converge. It also shows the formula's meaning. Over a tiny horizontal run $dx$ the curve rises $dy = f'(x)\,dx$ , and the little piece of curve is the hypotenuse $ds = \sqrt{dx^{2} + dy^{2}}$ of a right triangle. Factoring out $dx$ gives $ds = \sqrt{1 + (dy/dx)^2}\,dx$ , and $L = \int ds$ .

The symmetric form of the same fact is useful enough to record:

(ds)^2 = (dx)^2 + (dy)^2, \qquad L = \int_c^d\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy \text{ for a curve } x = g(y).

Definition 6.39 (Arc length function). For a smooth curve $y = f(x)$ starting at the point $(a, f(a))$ , the arc length function is

s(x) = \int_a^x\sqrt{1 + \bigl[f'(t)\bigr]^2}\,dt,

the distance along the curve from the starting point to $(x, f(x))$ . By the Fundamental Theorem of Calculus,

\frac{ds}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}.

The derivative $ds/dx$ is always at least $1$ : the curve is never shorter than its horizontal projection, and it equals $1$ exactly where the tangent is horizontal.

Now the bad news. The integrand $\sqrt{1 + (f')^2}$ has an elementary antiderivative only for a narrow family of functions. Even $y = x^{2}$ leads to $\int\sqrt{1 + 4x^{2}}\,dx$ , which needs a trigonometric substitution and produces a logarithm; $y = x^{3}$ and $y = \sin x$ lead to integrals with no elementary antiderivative at all, and those are handled numerically. Textbook exercises are therefore engineered so that $1 + (f')^2$ is a perfect square, and it pays to recognize the pattern. If

f'(x) = u(x) - \frac{1}{4u(x)} \qquad\text{then}\qquad 1 + [f'(x)]^2 = u^2 + \frac12 + \frac{1}{16u^2} = \left(u(x) + \frac{1}{4u(x)}\right)^2,

because the cross term $-\tfrac12$ in $(f')^2$ is exactly cancelled by the added $1$ . Functions like $\tfrac{x^{3}}{6} + \tfrac{1}{2x}$ , $\tfrac{x^{2}}{4} - \tfrac{\ln x}{2}$ and $x^{2} - \tfrac{\ln x}{8}$ are all of this type, and the square root disappears.

Intuition. Lay a piece of string along the curve, then pull it straight and measure it. Arc length is what the ruler reads. The formula computes the same thing by pretending the curve is made of very short straight segments, each the hypotenuse of a tiny right triangle with legs $dx$ and $dy$ : Pythagoras on each, then add. The factor $\sqrt{1 + (f')^2}$ is the ratio of hypotenuse to horizontal leg; a steep curve packs a lot of length into a little horizontal distance.

Example 6.40 (A curve with an elementary arc length). Find the length of the arc of $y = x^{3/2}$ from $(1,1)$ to $(4,8)$ .

Solution.

$f'(x) = \tfrac32 x^{1/2}$ , so $1 + [f'(x)]^2 = 1 + \tfrac94 x$ .
The length is

L = \int_1^4\sqrt{1 + \tfrac94 x}\,dx.

Substitute $u = 1 + \tfrac94 x$ , $du = \tfrac94\,dx$ ; when $x = 1$ , $u = \tfrac{13}{4}$ , and when $x = 4$ , $u = 10$ :

L = \frac49\int_{13/4}^{10}\sqrt{u}\,du = \frac49\cdot\frac23\Bigl[u^{3/2}\Bigr]_{13/4}^{10} = \frac{8}{27}\left(10\sqrt{10} - \frac{13\sqrt{13}}{8}\right) = \frac{80\sqrt{10} - 13\sqrt{13}}{27} \approx 7.63.

Sanity check: the straight segment from $(1,1)$ to $(4,8)$ has length $\sqrt{9 + 49} \approx 7.62$ ; the curve must be slightly longer, and it is.

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Example 6.41 (The perfect-square family). Find the length of the curve $y = \dfrac{x^{3}}{6} + \dfrac{1}{2x}$ for $1 \le x \le 2$ .

Solution.

$y' = \tfrac{x^{2}}{2} - \tfrac{1}{2x^{2}}$ . Squaring,

(y')^2 = \frac{x^4}{4} - \frac12 + \frac{1}{4x^4}.

Adding $1$ flips the sign of the middle term:

1 + (y')^2 = \frac{x^4}{4} + \frac12 + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2.

The square root is now the positive quantity $\tfrac{x^{2}}{2} + \tfrac{1}{2x^{2}}$ , so

L = \int_1^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 = \left(\frac{8}{6} - \frac14\right) - \left(\frac16 - \frac12\right) = \frac{13}{12} + \frac13 = \frac{17}{12}.

Sanity check: the endpoints are $(1, \tfrac23)$ and $(2, \tfrac{19}{12})$ , at straight-line distance $\sqrt{1 + (11/12)^2} \approx 1.36$ , and $\tfrac{17}{12} \approx 1.42$ is a little longer. Good.

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Example 6.42 (Arc length of a parabola). Find the length of $y = x^{2}$ from $x = 0$ to $x = 1$ .

Solution.

$y' = 2x$ , so $L = \int_0^1\sqrt{1 + 4x^{2}}\,dx$ . This is not a perfect square; substitute $u = 2x$ , $dx = \tfrac12 du$ :

L = \frac12\int_0^2\sqrt{1 + u^2}\,du.

The integral $\int\sqrt{1 + u^{2}}\,du$ is evaluated in the chapter on techniques of integration by the substitution $u = \tan\theta$ , which turns it into $\int\sec^{3}\theta\,d\theta$ . The result is

\int\sqrt{1 + u^2}\,du = \frac12\left(u\sqrt{1 + u^2} + \ln\bigl(u + \sqrt{1+u^2}\bigr)\right) + C.

Evaluate from $0$ to $2$ : the bracket at $u = 2$ is $2\sqrt5 + \ln(2 + \sqrt5)$ and at $u = 0$ it is $0$ . So

L = \frac12\cdot\frac12\left(2\sqrt5 + \ln(2 + \sqrt5)\right) = \frac{\sqrt5}{2} + \frac{\ln(2 + \sqrt5)}{4} \approx 1.479.

Sanity check: the chord from $(0,0)$ to $(1,1)$ has length $\sqrt2 \approx 1.414$ , and the path along the two legs has length $2$ . The arc is between them.

□

Example 6.43 (The arc length function). Find the arc length function for $y = x^{2} - \tfrac18\ln x$ measured from the point $(1,1)$ , and use it to find the length of the curve from $(1,1)$ to $(3, 9 - \tfrac18\ln 3)$ .

Solution.

$f'(x) = 2x - \tfrac{1}{8x}$ , which has the perfect-square shape with $u = 2x$ :

1 + [f'(x)]^2 = 4x^2 + \frac12 + \frac{1}{64x^2} = \left(2x + \frac{1}{8x}\right)^2.

Then

s(x) = \int_1^x\left(2t + \frac{1}{8t}\right)dt = \left[t^2 + \frac18\ln t\right]_1^x = x^2 + \frac18\ln x - 1.

The length to $x = 3$ is $s(3) = 9 + \tfrac18\ln 3 - 1 = 8 + \tfrac18\ln 3 \approx 8.14$ .

Once $s$ is known, the length between any two points on the curve is a difference of two values of $s$ ; this is the sense in which $s$ is a coordinate along the curve.□

Pitfall. The formula is $\sqrt{1 + (f')^2}$ , not $\sqrt{1 + f'}$ and not $1 + \sqrt{(f')^2}$ . And when the perfect-square trick produces $\sqrt{(u + \tfrac{1}{4u})^2}$ , the square root is $|u + \tfrac{1}{4u}|$ , which equals $u + \tfrac{1}{4u}$ only where that expression is positive. On the intervals used in this chapter it always is, but check it.

Exercises for §6.7 →

6.8Parametric curves: the formulas carry over

Everything in this chapter has a version for a curve given parametrically as $x = x(t)$ , $y = y(t)$ , $\alpha \le t \le \beta$ . The full treatment, including derivatives, tangent lines, areas and polar coordinates, is in the chapter on Parametric Equations and Polar Coordinates; this section records only what is needed to transfer the integrals.

Notation. For a smooth parametric curve, the slope of the tangent, the arc length element and the area under the curve are

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt, \qquad A = \int_\alpha^\beta y(t)\,x'(t)\,dt,

the last one requiring the curve to be traversed once, left to right, as $t$ increases. Arc length is $L = \int_\alpha^\beta ds$ and the area of the surface obtained by rotating the curve about the $x$ -axis is $\int_\alpha^\beta 2\pi\,y(t)\,ds$ .

The arc length element is the parametric form of $(ds)^2 = (dx)^2 + (dy)^2$ : divide by $(dt)^2$ and take the square root. The tangent slope is undefined where $dx/dt = 0$ (a vertical tangent, if $dy/dt \ne 0$ there) and zero where $dy/dt = 0$ .

Example 6.44 (Arc length of a parametric curve). Find the length of the curve $x = t^{2}$ , $y = t^{3}$ for $0 \le t \le 1$ .

Solution.

$dx/dt = 2t$ and $dy/dt = 3t^{2}$ , so $ds = \sqrt{4t^{2} + 9t^{4}}\,dt = t\sqrt{4 + 9t^{2}}\,dt$ (for $t \ge 0$ ).
Substitute $u = 4 + 9t^{2}$ , $du = 18t\,dt$ :

L = \int_0^1 t\sqrt{4 + 9t^2}\,dt = \frac{1}{18}\int_4^{13}\sqrt{u}\,du = \frac{1}{18}\cdot\frac23\Bigl[u^{3/2}\Bigr]_4^{13} = \frac{13\sqrt{13} - 8}{27} \approx 1.44.

Sanity check: the curve is $y = x^{3/2}$ from $(0,0)$ to $(1,1)$ , whose chord has length $\sqrt2 \approx 1.41$ .

□

Exercises for §6.8 →

6.9Area of a surface of revolution

Rotating a curve (rather than a region) about an axis sweeps out a surface, and its area is the amount of paint needed to coat it. The natural approximation is to replace the curve by a polygon, as for arc length, and rotate the polygon: each segment sweeps out a band, and each band is a piece of a cone.

A cone of base radius $r$ and slant height $\ell$ has lateral area $\pi r\ell$ (cut it along a slant line and flatten it into a sector of a circle of radius $\ell$ and arc length $2\pi r$ ; the sector's area is $\tfrac12\ell\cdot 2\pi r$ ). A frustum, the band cut from a cone between two parallel planes, with radii $r_1$ and $r_2$ and slant height $\ell$ , therefore has lateral area

A = \pi(r_1 + r_2)\,\ell = 2\pi r\ell, \qquad r = \frac{r_1 + r_2}{2},

obtained by subtracting two cone areas and using similar triangles. In words: a band's area is $2\pi$ times its average radius times its slant width.

Theorem 6.45 (Area of a surface of revolution). If $f \ge 0$ and $f$ has a continuous derivative on $[a,b]$ , the area of the surface obtained by rotating the curve $y = f(x)$ , $a \le x \le b$ , about the $x$ -axis is

S = \int_a^b 2\pi f(x)\sqrt{1 + \bigl[f'(x)\bigr]^2}\,dx.

Proof. Partition $[a,b]$ and inscribe the polygon with vertices $P_i = (x_i, f(x_i))$ . Rotating the segment $P_{i-1}P_i$ about the $x$ -axis produces a frustum with slant height $|P_{i-1}P_i|$ and average radius $\tfrac12\bigl(f(x_{i-1}) + f(x_i)\bigr)$ , so its area is

2\pi\cdot\frac{f(x_{i-1}) + f(x_i)}{2}\cdot|P_{i-1}P_i|.

As in the arc length proof, the Mean Value Theorem gives $|P_{i-1}P_i| = \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x$ for some $x_i^*$ in $[x_{i-1}, x_i]$ , and because $f$ is continuous and $\Delta x$ is small, both $f(x_{i-1})$ and $f(x_i)$ are close to $f(x_i^*)$ . The band's area is therefore approximately $2\pi f(x_i^*)\sqrt{1 + [f'(x_i^*)]^2}\,\Delta x$ , the sum over the bands is a Riemann sum for $2\pi f\sqrt{1 + (f')^2}$ , and its limit is the integral. Justifying the replacement of $f(x_{i-1})$ and $f(x_i)$ by $f(x_i^*)$ inside the limit needs one more fact. Let $\omega$ be the largest change in $f$ over any single subinterval of the partition and $M$ the maximum of $\sqrt{1 + (f')^2}$ on $[a,b]$ . The average of $f(x_{i-1})$ and $f(x_i)$ differs from $f(x_i^*)$ by at most $\omega$ , so the error in one band is at most $2\pi\,\omega\,M\,\Delta x$ and the total error over all bands is at most $2\pi\,\omega\,M\,(b - a)$ . That $\omega \to 0$ as the mesh goes to zero is the uniform continuity of $f$ on the closed interval $[a,b]$ , a real-analysis fact we take on trust; with it, the frustum sums and the Riemann sums have the same limit.∎

Remark. Where the curve meets the axis with a vertical tangent — the poles of the sphere below, or $y = \sqrt{x}$ at $x = 0$ — $f'$ is unbounded at an endpoint and the integral is, strictly, improper. In every such case in this chapter the factor $f$ vanishes fast enough to tame the root: the integrand simplifies to something bounded, and the improper integral is an ordinary one in disguise.

The compact way to remember the theorem is

S = \int 2\pi\,(\text{radius})\,ds, \qquad ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx = \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy,

where the radius is the distance from the curve to the axis. For rotation about the $x$ -axis the radius is $y$ ; for rotation about the $y$ -axis it is $x$ . Either arc length element may be used with either axis, so the four combinations are

S = \int 2\pi y\,ds \ \ (\text{about the } x\text{-axis}), \qquad S = \int 2\pi x\,ds \ \ (\text{about the } y\text{-axis}),

with $ds$ written in whichever variable makes the integral tractable.

The warning from arc length applies twice over: the factor $\sqrt{1 + (f')^2}$ is present, and now it is multiplied by $f$ . Exercises are chosen so that a substitution kills the root; sphere, cone and the perfect-square family are the standard sources of clean answers.

Intuition. Wrap masking tape around a vase in thin horizontal strips. Each strip is a ring whose length is the vase's circumference at that height, $2\pi r$ , and whose width is a short piece of the vase's profile — the slanted profile, not the vertical drop, because on a sloping surface the tape runs along the slope. Circumference times slant width, summed over all the strips, is the surface area.

Example 6.46 (Surface area of a sphere). Show that a sphere of radius $r$ has surface area $4\pi r^{2}$ .

Solution.

The sphere is generated by rotating the semicircle $y = \sqrt{r^{2} - x^{2}}$ , $-r \le x \le r$ , about the $x$ -axis.
$y' = -x/\sqrt{r^{2} - x^{2}}$ , so

1 + (y')^2 = 1 + \frac{x^2}{r^2 - x^2} = \frac{r^2}{r^2 - x^2}, \qquad \sqrt{1 + (y')^2} = \frac{r}{\sqrt{r^2 - x^2}}.

The square root in $y$ cancels the one in $ds$ :

S = \int_{-r}^{r}2\pi\sqrt{r^2 - x^2}\cdot\frac{r}{\sqrt{r^2 - x^2}}\,dx = 2\pi r\int_{-r}^{r}dx = 4\pi r^2.

The integrand is constant: every slab of the sphere of thickness $dx$ carries the same surface area $2\pi r\,dx$ , regardless of where it is cut. Archimedes knew this.□

Example 6.47 (Surface area of a cone). Find the area of the surface generated by rotating $y = x$ , $0 \le x \le 1$ , about the $x$ -axis.

Solution.

$y' = 1$ , so $\sqrt{1 + (y')^2} = \sqrt2$ .
Then

S = \int_0^1 2\pi x\sqrt2\,dx = 2\sqrt2\pi\left[\frac{x^2}{2}\right]_0^1 = \sqrt2\,\pi.

Sanity check: this is a cone of base radius $1$ and slant height $\sqrt2$ , and $\pi r\ell = \pi\sqrt2$ agrees.

□

Example 6.48 (Rotation about the $y$ -axis). The arc of the parabola $y = x^{2}$ from $(1,1)$ to $(2,4)$ is rotated about the $y$ -axis. Find the area of the resulting surface.

Solution.

The radius is $x$ and we may take $ds$ in the variable $x$ : with $y' = 2x$ , $ds = \sqrt{1 + 4x^{2}}\,dx$ .
So

S = \int_1^2 2\pi x\sqrt{1 + 4x^2}\,dx.

Substitute $u = 1 + 4x^{2}$ , $du = 8x\,dx$ ; the limits become $5$ and $17$ :

S = \frac{2\pi}{8}\int_5^{17}\sqrt{u}\,du = \frac{\pi}{4}\cdot\frac23\Bigl[u^{3/2}\Bigr]_5^{17} = \frac{\pi}{6}\left(17\sqrt{17} - 5\sqrt5\right) \approx 30.85.

Sanity check: the surface is a bowl between radii $1$ and $2$ with slant length about $3.17$ (the arc length of the parabola), so it is roughly a frustum of average radius $1.5$ : $2\pi(1.5)(3.17) \approx 29.9$ . Close.

□

Pitfall. Surface area needs $ds$ , not $dx$ . Forgetting the factor $\sqrt{1 + (f')^2}$ gives $\int 2\pi f\,dx$ , which is $2\pi$ times the area under the curve and has nothing to do with the surface. The two ideas are easy to confuse because the surface integrand contains the disk method's radius; the difference is that a surface has thickness $ds$ along the slope, while a disk has thickness $dx$ along the axis.

Exercises for §6.9 →

6.10Hydrostatic force

A fluid at rest presses on any surface in contact with it. At depth $d$ below the surface of a fluid of density $\rho$ , the pressure is the weight of the column of fluid above a unit area,

P = \rho g d,

measured in pascals (newtons per square meter) or pounds per square foot, and by Pascal's principle it acts equally in all directions, so a vertical plate feels it head-on. On a horizontal plate at constant depth the total force is simply pressure times area. On a vertical plate the depth changes from top to bottom, the pressure changes with it, and only an integral adds up the varying push.

Proposition 6.49 (Hydrostatic force on a vertical plate). Suppose a vertical plate is submerged in a fluid of density $\rho$ , and that at position $y$ (in some vertical coordinate running from $y = a$ to $y = b$ over the plate) the plate has horizontal width $w(y)$ and lies at depth $d(y)$ below the surface. The total force on one side of the plate is

F = \int_a^b\rho g\,d(y)\,w(y)\,dy.

The derivation is slicing again. Partition the plate into thin horizontal strips of height $\Delta y$ . Every point of a strip is at nearly the same depth $d(y_i^*)$ , so the pressure across the strip is nearly the constant $\rho g\,d(y_i^*)$ and the force on the strip is nearly pressure times area, $\rho g\,d(y_i^*)\cdot w(y_i^*)\,\Delta y$ . Sum and take the limit. Horizontal strips are compulsory; a vertical strip spans many depths and has no single pressure.

In US units the density is given as a weight density $\delta$ (for water, $62.5\ \text{lb/ft}^3$ ) and the formula reads $F = \int\delta\,d(y)\,w(y)\,dy$ ; in SI units $\rho g = 9800\ \text{N/m}^3$ for water. As with pumping, the two functions to get right are the width $w(y)$ , which comes from the plate's outline, and the depth $d(y)$ , which depends on where the surface is relative to your coordinate.

Intuition. Dive to the bottom of a swimming pool and your ears tell you the pressure is greater there than at the surface. A dam's face feels the same thing: the strip along its top is barely pushed, the strip along its bottom is pushed hard. To get the total push you cannot multiply one pressure by the whole area; you add the push on each thin horizontal strip, weighted by how deep it sits. This is why dams are thick at the base and thin at the crest.

Example 6.50 (Force on a rectangular gate). A rectangular gate $3$ m wide and $2$ m tall is submerged vertically in water so that its top edge is $4$ m below the surface. Find the hydrostatic force on the gate.

Solution.

Let $y$ be the distance below the top edge of the gate, from $0$ to $2$ . Then the depth is $d(y) = 4 + y$ and the width is the constant $w(y) = 3$ .
Integrate:

F = \int_0^2 9800\,(4 + y)\cdot 3\,dy = 29{,}400\left[4y + \frac{y^2}{2}\right]_0^2 = 29{,}400\cdot 10 = 294{,}000\ \text{N}.

Sanity check: the pressure at the center of the gate (depth $5$ m) is $49{,}000$ Pa, and the gate's area is $6\ \text{m}^2$ ; their product is $294{,}000$ N. For a rectangle, pressure at the center times area is exact, because the pressure varies linearly with depth.

□

Example 6.51 (Force on a semicircular plate). A plate in the shape of a semicircle of radius $2$ m is submerged vertically in water with its diameter along the surface and the curved edge below. Find the force on one side.

Solution.

Let $y$ be the depth below the surface, $0 \le y \le 2$ . Then $d(y) = y$ . At depth $y$ the plate's horizontal width is the chord of the circle $x^{2} + y^{2} = 4$ , so $w(y) = 2\sqrt{4 - y^{2}}$ .
The force is

F = \int_0^2 9800\,y\cdot 2\sqrt{4 - y^2}\,dy.

Substitute $u = 4 - y^{2}$ , $du = -2y\,dy$ , so $2y\,dy = -du$ ; the limits $y = 0, 2$ become $u = 4, 0$ :

F = 9800\int_0^4\sqrt{u}\,du = 9800\cdot\frac23\cdot 8 = \frac{156{,}800}{3} \approx 52{,}300\ \text{N}.

Sanity check: the plate's area is $2\pi \approx 6.28\ \text{m}^2$ and its centroid (computed in the next section) is at depth $\tfrac{4\cdot 2}{3\pi} \approx 0.85$ m. Pressure at the centroid times area is $9800 \times 0.85 \times 6.28 \approx 52{,}300$ N. Same answer.

□

Example 6.52 (Force on a triangular plate below the surface). A vertical plate is an isosceles triangle with its $4$ m base horizontal at the top and its apex $3$ m directly below the midpoint of the base. The base is $1$ m below the surface of the water. Find the force on one side.

Solution.

Let $y$ be the distance below the base, $0 \le y \le 3$ . The depth is $d(y) = 1 + y$ .
The width shrinks linearly from $4$ at $y = 0$ to $0$ at $y = 3$ : by similar triangles $w(y) = 4\left(1 - \tfrac{y}{3}\right)$ .
Assemble and expand:

F = \int_0^3 9800\,(1 + y)\cdot 4\left(1 - \frac{y}{3}\right)dy = 39{,}200\int_0^3\left(1 + \frac{2y}{3} - \frac{y^2}{3}\right)dy = 39{,}200\left[y + \frac{y^2}{3} - \frac{y^3}{9}\right]_0^3.

The bracket is $3 + 3 - 3 = 3$ , so $F = 117{,}600$ N.

Sanity check: the triangle's area is $6\ \text{m}^2$ and its centroid is one third of the way from the base to the apex, at depth $1 + 1 = 2$ m; $9800 \times 2 \times 6 = 117{,}600$ N.□

The sanity checks in the last three examples all used the same shortcut, which we can now state: the hydrostatic force on a vertical plate equals the pressure at the plate's centroid times the plate's area,

F = \rho g\,\bar d\,A.

It follows from the definition of the centroid in the next section, since $\int d(y)\,w(y)\,dy$ is the first moment of the plate about the surface line, which is $\bar d\,A$ . The shortcut is exact, and it is the fastest check on any hydrostatic calculation whose shape has a known centroid.

Pitfall. Depth is measured down from the fluid surface, not up from the bottom of the plate and not from the top of the plate unless the top is at the surface. If a plate's top edge is $4$ m under water, a strip $y$ meters below that edge is at depth $4 + y$ . Getting this offset wrong changes the answer by a factor, not by a rounding error.

Exercises for §6.10 →

6.11Moments and centers of mass

Where should you place a fulcrum under a plank so that it balances? Where is the point at which a flat metal plate can be balanced on a pin? The answer is the center of mass, and its coordinates are ratios of integrals.

Start with point masses $m_1, \dots, m_n$ on a line at positions $x_1, \dots, x_n$ . The rod balances at $\bar x$ when the turning effects on the two sides cancel, $\sum m_i(x_i - \bar x) = 0$ , that is, when

\bar x = \frac{\sum m_i x_i}{\sum m_i} = \frac{M}{m},

the total moment about the origin divided by the total mass. The center of mass is the mass-weighted average position.

Definition 6.53 (Center of mass of a rod). For a thin rod lying along the $x$ -axis from $a$ to $b$ with linear density $\rho(x)$ (mass per unit length), the mass and the moment about the origin are

m = \int_a^b\rho(x)\,dx, \qquad M = \int_a^b x\,\rho(x)\,dx,

and the center of mass is $\bar x = M/m$ .

Slice the rod: the piece over $[x_{i-1}, x_i]$ has mass approximately $\rho(x_i^*)\,\Delta x$ located at $x_i^*$ , so its moment is $x_i^*\rho(x_i^*)\,\Delta x$ . Summing and passing to the limit gives the two integrals. When $\rho$ is constant it cancels from the ratio and $\bar x$ is the midpoint; variable density pulls $\bar x$ toward the heavy end.

For a flat plate (a lamina) occupying a region in the plane, balance must hold in two directions. The moment about the $y$ -axis, $M_y$ , measures the tendency to rotate about that axis and is built from $x$ -coordinates; the moment about the $x$ -axis, $M_x$ , is built from $y$ -coordinates. For point masses, $M_y = \sum m_i x_i$ and $M_x = \sum m_i y_i$ , and $(\bar x, \bar y) = (M_y/m, M_x/m)$ .

Theorem 6.54 (Centroid of a plane region). Let $\mathcal R$ be the region under $y = f(x)$ , above the $x$ -axis, from $x = a$ to $x = b$ , with $f$ continuous and $f \ge 0$ , and let $A = \int_a^b f(x)\,dx$ be its area. If a lamina of uniform density $\rho$ occupies $\mathcal R$ , its center of mass, called the centroid of $\mathcal R$ , is $(\bar x, \bar y)$ with

\bar x = \frac{1}{A}\int_a^b x\,f(x)\,dx, \qquad \bar y = \frac{1}{A}\int_a^b\frac12\bigl[f(x)\bigr]^2\,dx.

Proof. Slice $\mathcal R$ into vertical strips. The strip over $[x_{i-1}, x_i]$ is nearly a rectangle of width $\Delta x$ and height $f(x_i^*)$ , where we take $x_i^*$ to be the midpoint. Its mass is $\rho f(x_i^*)\,\Delta x$ , and by symmetry a rectangle balances at its center, the point $\bigl(x_i^*, \tfrac12 f(x_i^*)\bigr)$ . Treating the strip as a point mass at its center, its moment about the $y$ -axis is $x_i^*\cdot\rho f(x_i^*)\,\Delta x$ and its moment about the $x$ -axis is $\tfrac12 f(x_i^*)\cdot\rho f(x_i^*)\,\Delta x$ . Summing over strips and taking the limit,

M_y = \rho\int_a^b x\,f(x)\,dx, \qquad M_x = \rho\int_a^b\frac12\bigl[f(x)\bigr]^2\,dx, \qquad m = \rho A.

Dividing, the density cancels and the formulas for $\bar x = M_y/m$ and $\bar y = M_x/m$ follow.∎

The density canceled, which is why the centroid is a property of the region's shape alone; for a lamina of non-uniform density the same slicing works but $\rho$ stays inside the integrals (and, if $\rho$ depends on $y$ , the strips must be cut the other way). For a region between two curves $f(x) \ge g(x)$ the strip runs from $g$ to $f$ , its center is at height $\tfrac12(f + g)$ , and the same argument gives

\bar x = \frac{1}{A}\int_a^b x\bigl[f(x) - g(x)\bigr]dx, \qquad \bar y = \frac{1}{A}\int_a^b\frac12\left(\bigl[f(x)\bigr]^2 - \bigl[g(x)\bigr]^2\right)dx,

since $\tfrac12(f + g)(f - g) = \tfrac12(f^{2} - g^{2})$ .

Proposition 6.55 (Symmetry principle). If a region is symmetric about a line, its centroid lies on that line.

Reflection in the line pairs each strip with a mirror strip of equal mass on the other side, so the moments about the line cancel. The principle saves an integral every time it applies: a semicircle centered on the $y$ -axis has $\bar x = 0$ without calculation, and a region symmetric about two lines has its centroid at their intersection.

Theorem 6.56 (Theorem of Pappus). Let $\mathcal R$ be a plane region of area $A$ that does not cross a line $\ell$ in its plane (it lies on one side of $\ell$ , and may touch it), and let $d$ be the distance from the centroid of $\mathcal R$ to $\ell$ . The volume of the solid obtained by rotating $\mathcal R$ about $\ell$ is

V = 2\pi d\,A,

the area of the region times the distance traveled by its centroid.

Proof. Take $\ell$ to be the $y$ -axis and $\mathcal R$ the region between $f \ge g$ on $[a,b]$ with $a \ge 0$ . By the shell method,

V = \int_a^b 2\pi x\bigl[f(x) - g(x)\bigr]dx = 2\pi\int_a^b x\bigl[f(x) - g(x)\bigr]dx = 2\pi\,\bar x A,

by the centroid formula, and $\bar x = d$ is the distance from the centroid to the axis.∎

Pappus turns a hard volume into an easy one whenever the centroid is known, and it runs backward too: a known volume and area give the centroid.

Intuition. Balance a ruler on your finger. With a lump of clay stuck near one end, the balance point shifts toward the clay: the heavier side gets more say in where the average lands. A flat plate is the same problem in two directions at once, and Pappus's theorem says that when you spin the plate around an axis, the plate's whole area is effectively carried around the circle its balance point traces, so volume is area times that circle's circumference.

Example 6.57 (Center of mass of a non-uniform rod). A rod extends from $x = 0$ to $x = 4$ meters and its density is $\rho(x) = 3x$ kg/m. Find its center of mass.

Solution.

Mass: $m = \int_0^4 3x\,dx = \left[\tfrac32 x^{2}\right]_0^4 = 24$ kg.
Moment: $M = \int_0^4 x\cdot 3x\,dx = \left[x^{3}\right]_0^4 = 64$ kg·m.
$\bar x = 64/24 = \tfrac83 \approx 2.67$ m.

The density increases to the right, so the balance point is right of the midpoint $2$ , as it should be. It cannot be beyond $4$ , and $\tfrac83$ is comfortably inside.□

Example 6.58 (Centroid of a semicircle). Find the centroid of the semicircular region $x^{2} + y^{2} \le r^{2}$ , $y \ge 0$ .

Solution.

The region is symmetric about the $y$ -axis, so $\bar x = 0$ by the symmetry principle.
The upper boundary is $f(x) = \sqrt{r^{2} - x^{2}}$ on $[-r, r]$ and the area is $A = \tfrac12\pi r^{2}$ . Then

\bar y = \frac{1}{A}\int_{-r}^{r}\frac12\left(r^2 - x^2\right)dx = \frac{2}{\pi r^2}\cdot\frac12\cdot 2\int_0^r\left(r^2 - x^2\right)dx = \frac{2}{\pi r^2}\cdot\frac{2r^3}{3} = \frac{4r}{3\pi}.

Sanity check: $\tfrac{4}{3\pi} \approx 0.42$ , so the centroid is a bit less than halfway up. There is more area near the diameter than near the top of the arc, which pulls the balance point down. Correct.

□

Example 6.59 (Centroid of the region under a cosine arch). Find the centroid of the region bounded by $y = \cos x$ , $y = 0$ , $x = 0$ and $x = \pi/2$ .

Solution.

Area: $A = \int_0^{\pi/2}\cos x\,dx = 1$ .
$\bar x = \int_0^{\pi/2} x\cos x\,dx$ . Integrate by parts with $u = x$ , $dv = \cos x\,dx$ :

\bar x = \Bigl[x\sin x\Bigr]_0^{\pi/2} - \int_0^{\pi/2}\sin x\,dx = \frac{\pi}{2} - 1 \approx 0.571.

$\bar y = \int_0^{\pi/2}\tfrac12\cos^{2} x\,dx = \tfrac14\int_0^{\pi/2}(1 + \cos 2x)\,dx = \tfrac14\left[x + \tfrac12\sin 2x\right]_0^{\pi/2} = \tfrac{\pi}{8} \approx 0.393$ .
Sanity check: the region is fat on the left (where $\cos x \approx 1$ ) and thin on the right, so $\bar x$ should be less than the midpoint $\pi/4 \approx 0.785$ ; and $\bar y$ should be under $\tfrac12$ since most of the area is low. Both hold.

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Example 6.60 (Centroid of a region between two curves). Find the centroid of the region bounded by $y = x$ and $y = x^{2}$ .

Solution.

The curves meet at $0$ and $1$ , the line is on top, and $A = \int_0^1(x - x^{2})\,dx = \tfrac12 - \tfrac13 = \tfrac16$ .
$\bar x = 6\int_0^1 x(x - x^{2})\,dx = 6\left(\tfrac13 - \tfrac14\right) = \tfrac12$ .
$\bar y = 6\int_0^1\tfrac12\left(x^{2} - x^{4}\right)dx = 3\left(\tfrac13 - \tfrac15\right) = \tfrac25$ .

Sanity check: the region is the sliver between a line and a parabola, thickest around $x = \tfrac12$ , and $\bar x = \tfrac12$ is exactly where you would put a pin by eye. And $(\tfrac12, \tfrac25)$ lies inside the region, since $\tfrac14 < \tfrac25 < \tfrac12$ .□

Example 6.61 (Volume of a torus by Pappus). A circle of radius $r$ is rotated about a line in its plane at distance $R > r$ from its center. Find the volume of the resulting torus (a doughnut).

Solution.

The disk has area $A = \pi r^{2}$ , and its centroid, by symmetry, is its center, at distance $d = R$ from the axis.
Pappus:

V = 2\pi d\,A = 2\pi R\cdot\pi r^2 = 2\pi^2 r^2 R.

As a second check on Pappus, rotate the semicircular region of radius $r$ about its diameter: $A = \tfrac12\pi r^{2}$ and $d = \tfrac{4r}{3\pi}$ from the semicircle example, so $V = 2\pi\cdot\tfrac{4r}{3\pi}\cdot\tfrac12\pi r^{2} = \tfrac43\pi r^{3}$ , the volume of a sphere.

Doing the torus with washers means integrating $\sqrt{r^{2} - y^{2}}$ terms; Pappus makes it a one-line product.□

Pitfall. $M_x$ is the moment about the $x$ -axis and it is built from $y$ -coordinates, so $\bar y = M_x/m$ , not $M_x$ itself and not $M_y/m$ . The letter in the subscript names the axis, not the coordinate. And in $\bar y$ the integrand is $\tfrac12 f^{2}$ , because a strip's mass sits at half its height; forgetting the $\tfrac12$ doubles $\bar y$ and usually puts the "centroid" outside the region, which is an immediate signal that something is wrong.

Exercises for §6.11 →

6.12Applications to economics and biology

The slicing argument does not care whether the thing being sliced is a solid, a fluid or a market. Three short applications show the range.

Definition 6.62 (Consumer surplus). Let $p(x)$ be the demand function for a commodity: the price at which exactly $x$ units will be sold. If the commodity sells at the price $P = p(X)$ corresponding to sales of $X$ units, the consumer surplus is

\text{CS} = \int_0^X\bigl[p(x) - P\bigr]\,dx.

The demand curve is decreasing: the first units are bought by customers who would have paid a lot, and later units by customers who will only pay a little. If the market price is $P$ , the customer who would have paid $p(x)$ for the $x$ -th unit saves $p(x) - P$ . Partition $[0, X]$ into groups of $\Delta x$ units; the group near $x$ saves about $[p(x_i^*) - P]\,\Delta x$ , and the total savings across all buyers is the Riemann sum, hence the integral. Geometrically it is the area between the demand curve and the horizontal line $y = P$ from $0$ to $X$ .

Example 6.63 (Consumer surplus). The demand for a product is $p(x) = 1200 - 0.2x - 0.0001x^{2}$ dollars per unit. Find the consumer surplus when the sales level is $500$ units.

Solution.

The price at $X = 500$ is $P = 1200 - 100 - 25 = 1075$ .
Then

\text{CS} = \int_0^{500}\left(1200 - 0.2x - 0.0001x^2 - 1075\right)dx = \int_0^{500}\left(125 - 0.2x - 0.0001x^2\right)dx.

Evaluate: $\left[125x - 0.1x^{2} - \tfrac{0.0001}{3}x^{3}\right]_0^{500} = 62{,}500 - 25{,}000 - \tfrac{12{,}500}{3} \approx 33{,}333$ dollars.

□

Theorem 6.64 (Poiseuille's law of laminar flow). In a blood vessel (or pipe) of radius $R$ and length $l$ , with pressure difference $P$ between the ends and viscosity $\eta$ , the velocity of the fluid at distance $r$ from the central axis is $v(r) = \dfrac{P}{4\eta l}\left(R^{2} - r^{2}\right)$ , and the flux, the volume passing a cross-section per unit time, is

F = \frac{\pi P R^4}{8\eta l}.

Proof. The velocity law is a fact from fluid mechanics; the flux follows from it by slicing the cross-section into thin concentric rings. The ring between radii $r$ and $r + \Delta r$ has area approximately $2\pi r\,\Delta r$ , and the fluid in it moves at velocity $v(r)$ , so the volume crossing that ring per unit time is about $v(r)\cdot 2\pi r\,\Delta r$ . Summing and taking the limit,

F = \int_0^R 2\pi r\,v(r)\,dr = \frac{2\pi P}{4\eta l}\int_0^R\left(R^2 r - r^3\right)dr = \frac{\pi P}{2\eta l}\left(\frac{R^4}{2} - \frac{R^4}{4}\right) = \frac{\pi P R^4}{8\eta l}.

∎

The fourth power is the physiologically important part: halving a vessel's radius cuts the flow by a factor of sixteen, which is why small changes in arterial radius have such large effects on circulation.

The last application is a measurement technique. In the dye dilution method for cardiac output, an amount $A$ of dye is injected into the right atrium and its concentration $c(t)$ in the aorta is measured over an interval $[0,T]$ until the dye has passed. If the heart pumps at a constant rate $F$ (volume per unit time), then in a short interval $\Delta t$ the volume $F\,\Delta t$ of blood carries $c(t)\,F\,\Delta t$ of dye, so the total dye passing the sensor is $\int_0^T c(t)F\,dt = A$ , and

F = \frac{A}{\int_0^T c(t)\,dt}.

Example 6.65 (Cardiac output). A $6$ mg bolus of dye is injected and its concentration in the aorta is modeled by $c(t) = 0.3\,t(10 - t)$ mg/L for $0 \le t \le 10$ seconds. Estimate the cardiac output.

Solution.

$\int_0^{10} 0.3\,(10t - t^{2})\,dt = 0.3\left[5t^{2} - \tfrac{t^{3}}{3}\right]_0^{10} = 0.3\left(500 - \tfrac{1000}{3}\right) = 0.3\cdot\tfrac{500}{3} = 50$ mg·s/L.
$F = 6/50 = 0.12$ L/s, which is $7.2$ L/min — a normal resting cardiac output.

□

Exercises for §6.12 →

6.13Probability

A continuous random variable $X$ — the height of a randomly chosen adult, the time until the next bus, the lifetime of a light bulb — does not have a probability of taking any single value; it has a probability of landing in an interval, and that probability is an integral.

Definition 6.66 (Probability density function). A function $f$ is a probability density function for the random variable $X$ if

P(a \le X \le b) = \int_a^b f(x)\,dx \quad\text{for all } a \le b, \qquad f(x) \ge 0 \text{ for all } x, \qquad \int_{-\infty}^{\infty}f(x)\,dx = 1.

The three conditions say that probability is area under the density, that areas are never negative, and that the total probability is $1$ . Note what the definition does not say: $f(x)$ is not the probability that $X = x$ (that probability is $\int_x^x f = 0$ ), and $f$ may exceed $1$ on a short interval as long as the total area is $1$ . The improper integral is the one studied in the chapter on techniques of integration; for a density that is zero outside a bounded interval it is an ordinary integral over that interval.

Definition 6.67 (Mean and median). The mean (or expected value) of a random variable with density $f$ is

\mu = \int_{-\infty}^{\infty}x\,f(x)\,dx,

and the median is the number $m$ such that $\int_{-\infty}^{m}f(x)\,dx = \tfrac12$ .

The mean is the probability-weighted average of the possible values, exactly the formula for the center of mass of a rod with density $f$ and total mass $1$ ; if the region under the density were cut from sheet metal, it would balance at $x = \mu$ . The median is the point that splits the area in half. For a symmetric density the two coincide; for a skewed one they do not, and the mean is dragged toward the long tail.

Theorem 6.68 (Exponential distribution). For $\lambda > 0$ , the function

f(t) = \begin{cases}0, & t < 0 \\ \lambda e^{-\lambda t}, & t \ge 0\end{cases}

is a probability density function, with mean $\mu = 1/\lambda$ , median $(\ln 2)/\lambda$ , and

P(X \le t) = 1 - e^{-\lambda t}, \qquad P(X > t) = e^{-\lambda t} \qquad (t \ge 0).

Proof. $f \ge 0$ is clear. For the total probability and the tail,

\int_0^{\infty}\lambda e^{-\lambda t}\,dt = \lim_{b\to\infty}\Bigl[-e^{-\lambda t}\Bigr]_0^b = \lim_{b\to\infty}\left(1 - e^{-\lambda b}\right) = 1, \qquad P(X > t) = \int_t^{\infty}\lambda e^{-\lambda s}\,ds = e^{-\lambda t}.

For the mean, integrate by parts with $u = t$ and $dv = \lambda e^{-\lambda t}\,dt$ , so $v = -e^{-\lambda t}$ :

\mu = \int_0^{\infty}t\,\lambda e^{-\lambda t}\,dt = \lim_{b\to\infty}\left(\Bigl[-te^{-\lambda t}\Bigr]_0^b + \int_0^b e^{-\lambda t}\,dt\right) = 0 + \frac{1}{\lambda},

using $be^{-\lambda b} \to 0$ (the exponential beats the polynomial). Finally the median solves $1 - e^{-\lambda m} = \tfrac12$ , giving $m = (\ln 2)/\lambda$ .∎

The exponential density models waiting times: the time until the next phone call, the next radioactive decay, the next customer. Its parameter is fixed by the mean: a process whose events are on average $\mu$ apart has $\lambda = 1/\mu$ . The median is always shorter than the mean, by the factor $\ln 2 \approx 0.69$ , because the density is skewed toward long waits.

Definition 6.69 (Normal distribution). The normal distribution with mean $\mu$ and standard deviation $\sigma > 0$ has density

f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-(x-\mu)^2/(2\sigma^2)}.

This is the bell curve, and it describes an enormous range of measured quantities: heights, test scores, measurement errors, anything that is the sum of many small independent effects. The graph is symmetric about $x = \mu$ , so the mean and median are both $\mu$ , and $\sigma$ controls the spread: small $\sigma$ gives a tall narrow peak, large $\sigma$ a low wide one. That $f$ integrates to $1$ depends on the classical fact

\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi},

which is proved with a double integral in the chapter on multiple integrals.

The normal density has no elementary antiderivative, so probabilities under it cannot be found by the Fundamental Theorem of Calculus. They are computed numerically (Simpson's rule, or a table of the standard normal $\mu = 0$ , $\sigma = 1$ , to which every case reduces by the substitution $z = (x - \mu)/\sigma$ ). Three values are worth memorizing:

P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.683, \qquad P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.954, \qquad P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.997.

Intuition. A density function is a histogram with infinitely thin bars. The height at a point is not a probability; the area over an interval is. Asking "what is the probability that the bus comes in exactly $4.000\ldots$ minutes" gets the answer zero, but "between $3$ and $5$ minutes" has an honest area. The mean is where the histogram would balance, and for a lopsided histogram like the exponential the balance point sits to the right of the halfway-area point, because the long tail, though thin, is far away and exerts leverage.

Example 6.70 (Checking a density and computing a probability). Let $f(x) = \tfrac{3}{32}x(4 - x)$ for $0 \le x \le 4$ and $f(x) = 0$ otherwise. Verify that $f$ is a probability density function, find $P(1 \le X \le 2)$ , and find the mean.

Solution.

$f \ge 0$ on $[0,4]$ since both $x$ and $4 - x$ are non-negative there. Total area:

\int_0^4\frac{3}{32}\left(4x - x^2\right)dx = \frac{3}{32}\left[2x^2 - \frac{x^3}{3}\right]_0^4 = \frac{3}{32}\left(32 - \frac{64}{3}\right) = \frac{3}{32}\cdot\frac{32}{3} = 1.

Probability:

P(1 \le X \le 2) = \frac{3}{32}\left[2x^2 - \frac{x^3}{3}\right]_1^2 = \frac{3}{32}\left[\left(8 - \frac83\right) - \left(2 - \frac13\right)\right] = \frac{3}{32}\cdot\frac{11}{3} = \frac{11}{32}.

Mean: the density is symmetric about $x = 2$ , so $\mu = 2$ by symmetry. Directly,

\mu = \frac{3}{32}\int_0^4\left(4x^2 - x^3\right)dx = \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4 = \frac{3}{32}\left(\frac{256}{3} - 64\right) = \frac{3}{32}\cdot\frac{64}{3} = 2.

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Example 6.71 (Exponential waiting times). Calls to a helpline arrive at random with an average gap of $4$ minutes between calls. Find the probability that the next call arrives within $2$ minutes, the probability that it takes more than $5$ minutes, and the median waiting time.

Solution.

The mean is $4$ , so $\lambda = \tfrac14$ and the density is $f(t) = \tfrac14 e^{-t/4}$ for $t \ge 0$ .
$P(X \le 2) = 1 - e^{-2/4} = 1 - e^{-1/2} \approx 0.393$ .
$P(X > 5) = e^{-5/4} \approx 0.287$ .
Median: $m = 4\ln 2 \approx 2.77$ minutes.

Note how the median ( $2.77$ ) is well below the mean ( $4$ ): more than half of all waits are shorter than average, balanced by a minority of long waits.□

Example 6.72 (A normal probability). Adult heights in a population are normally distributed with mean $175$ cm and standard deviation $7$ cm. What proportion of adults are between $168$ cm and $189$ cm tall?

Solution.

The interval runs from $\mu - \sigma$ to $\mu + 2\sigma$ .
By symmetry, $P(\mu - \sigma \le X \le \mu) = \tfrac12(0.683) \approx 0.341$ and $P(\mu \le X \le \mu + 2\sigma) = \tfrac12(0.954) \approx 0.477$ .
Adding, the proportion is approximately $0.341 + 0.477 = 0.818$ , about $82\%$ .

Written as an integral, this is

\int_{168}^{189}\frac{1}{7\sqrt{2\pi}}\,e^{-(x - 175)^2/98}\,dx,

and there is no antiderivative to evaluate it with; the tabulated values are the only route.□

Pitfall. A density can exceed $1$ ( $f(t) = 5e^{-5t}$ starts at $5$ ) and the probability of any exact value is $0$ . Both are unsettling on first meeting and both are correct: probability is area, and area over a single point is zero however tall the curve.

Summary. Every formula in this chapter is a Riemann sum for one slice. With $f,g$ continuous and $f\ge g$ on $[a,b]$ , $A=\int_a^b[f(x)-g(x)]\,dx$ (split at crossings; use $dy$ and right minus left when the region is described by $x=h(y)$ ). If the cross-section at $x$ has continuous area $A(x)$ , then $V=\int_a^b A(x)\,dx$ , which specialises to disks $V=\int_a^b\pi[f(x)]^2dx$ , washers $V=\int_a^b\pi([f]^2-[g]^2)dx$ , and shells $V=\int_a^b 2\pi x f(x)\,dx$ about the $y$ -axis. Work is $W=\int_a^b F(x)\,dx$ , with $F=kx$ for a spring and $W=\int\rho g\,A(y)\,d(y)\,dy$ for pumping; hydrostatic force is $F=\int_a^b\rho g\,d(y)w(y)\,dy$ over horizontal strips. The average value $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f$ is attained at some $c$ when $f$ is continuous. For $f'$ continuous, $L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx$ and, if also $f\ge 0$ , $S=\int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}\,dx$ . The centroid of the region under $f\ge 0$ is $\bar x=\frac1A\int x f\,dx$ , $\bar y=\frac1A\int\frac12 f^{2}\,dx$ , and Pappus gives $V=2\pi d A$ . A density satisfies $f\ge 0$ and $\int_{-\infty}^{\infty}f=1$ , with $\mu=\int_{-\infty}^{\infty}x f(x)\,dx$ .

Exercises for §6.13 →

Common Mistakes

Integrating $f - g$ across a crossing point. Solve $f = g$ first, and take top minus bottom on each piece; a single integral across a crossing measures signed area and can even be zero.
Choosing the integration variable by habit. If the left and right boundaries are single curves and the top or bottom switches formula, integrate in $y$ .
Forgetting to square the radius in disks and washers, or squaring the difference instead of subtracting the squares: a washer's area is $\pi(R^{2} - r^{2})$ , never $\pi(R - r)^2$ .
Measuring radii from the $x$ -axis when the axis of rotation is $y = c$ . The radius is $|f(x) - c|$ , and the curve farther from the axis is the outer one, whichever is higher.
Using $x$ as the shell radius for every vertical axis. For the axis $x = c$ the radius is the distance $|x - c|$ .
In pumping problems, measuring the lift to the surface of the water instead of to the outlet, or integrating over the whole tank when it is only partly full.
In spring problems, using the total length of the spring instead of the extension beyond the natural length.
Dropping the square root in arc length: it is $\sqrt{1 + (f')^2}$ , and the perfect-square trick applies only when $1 + (f')^2$ genuinely factors.
Using $dx$ instead of $ds$ in surface area, which computes $2\pi$ times an area under a curve rather than a surface.
Measuring depth from the wrong reference in hydrostatic problems. Depth is distance below the fluid surface; a plate whose top is $4$ m down has a strip $y$ below its top at depth $4 + y$ .
Confusing $M_x$ with $\bar x$ . $M_x$ is the moment about the $x$ -axis and is built from $y$ ; the coordinate is $\bar y = M_x/m$ . And $\bar y$ carries the factor $\tfrac12$ because a strip's mass sits at half its height.
Applying Pappus to a region that straddles the axis; the theorem needs the region on one side of it.
Reading a density value $f(x)$ as a probability. Probability is area, $\int_a^b f$ , and $f$ itself may exceed $1$ .
Not sketching. Nearly every error in this chapter — wrong limits, wrong outer radius, wrong depth — is caught by a rough picture with the axis drawn in.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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